The Decomposition Group

Suppose $ K$ is a number field that is Galois over $ \mathbf{Q}$ with group $ G=\Gal (K/\mathbf{Q})$. Fix a prime $ \mathfrak{p}\subset \O _K$ lying over $ p\in\mathbf{Z}$.

Definition 14.1.1 (Decomposition group)   The of $ \mathfrak{p}$ is the subgroup

$\displaystyle D_\mathfrak{p}= \{\sigma \in G : \sigma(\mathfrak{p})=\mathfrak{p}\} \leq G.

(Note: The decomposition group is called the ``splitting group'' in Swinnerton-Dyer. Everybody I know calls it the decomposition group, so we will too.)

Let $ \mathbf{F}_{\mathfrak{p}} = \O _K/\mathfrak{p}$ denote the residue class field of $ \mathfrak{p}$. In this section we will prove that there is a natural exact sequence

$\displaystyle 1\to I_\mathfrak{p}\to D_\mathfrak{p}\to \Gal (\mathbf{F}_{\mathfrak{p}}/\mathbf{F}_p)\to 1,

where $ I_\mathfrak{p}$ is the of $ D_\mathfrak{p}$, and $ \char93 I_\mathfrak{p}=e$. The most interesting part of the proof is showing that the natural map $ D_\mathfrak{p}\to \Gal (\mathbf{F}_{\mathfrak{p}}/\mathbf{F}_p)$ is surjective.

We will also discuss the structure of $ D_\mathfrak{p}$ and introduce Frobenius elements, which play a crucial roll in understanding Galois representations.

Recall that $ G$ acts on the set of primes $ \mathfrak{p}$ lying over $ p$. Thus the decomposition group is the stabilizer in $ G$ of $ \mathfrak{p}$. The orbit-stabilizer theorem implies that $ [G:D_\mathfrak{p}]$ equals the orbit of $ \mathfrak{p}$, which by Theorem 13.2.2 equals the number $ g$ of primes lying over $ p$, so $ [G:D_\mathfrak{p}]=g$.

Lemma 14.1.2   The decomposition subgroups $ D_\mathfrak{p}$ corresponding to primes $ \mathfrak{p}$ lying over a given $ p$ are all conjugate in $ G$.

Proof. We have $ \tau(\sigma(\tau^{-1}(\mathfrak{p}))) = \mathfrak{p}$ if and only if $ \sigma(\tau^{-1}(\mathfrak{p})) = \tau^{-1}\mathfrak{p}$. Thus $ \tau\sigma\tau^{-1}\in D_p$ if and only if $ \sigma\in D_{\tau^{-1}\mathfrak{p}}$, so $ \tau^{-1}D_p\tau = D_{\tau^{-1}\mathfrak{p}}$. The lemma now follows because, by Theorem 13.2.2, $ G$ acts transitively on the set of $ \mathfrak{p}$ lying over $ p$. $ \qedsymbol$

The decomposition group is extremely useful because it allows us to see the extension $ K/\mathbf{Q}$ as a tower of extensions, such that at each step in the tower we understand well the splitting behavior of the primes lying over $ p$. Now might be a good time to glance ahead at Figure 14.1.2 on page [*].

We characterize the fixed field of $ D=D_\mathfrak{p}$ as follows.

Proposition 14.1.3   The fixed field $ K^D$ of $ D$

$\displaystyle K^D=\{a \in K : \sigma(a) = a$ for all $\displaystyle \sigma \in D\}$

is the smallest subfield $ L\subset K$ such that $ \mathfrak{p}\cap L$ does not split in $ K$ (i.e., $ g(K/L)=1$).

Proof. First suppose $ L=K^D$, and note that by Galois theory $ \Gal (K/L)\cong
D$, and by Theorem 13.2.2, the group $ D$ acts transitively on the primes of $ K$ lying over $ \mathfrak{p}\cap L$. One of these primes is $ \mathfrak{p}$, and $ D$ fixes $ \mathfrak{p}$ by definition, so there is only one prime of $ K$ lying over $ \mathfrak{p}\cap L$, i.e., $ \mathfrak{p}\cap L$ does not split in $ K$. Conversely, if $ L\subset K$ is such that $ \mathfrak{p}\cap L$ does not split in $ K$, then $ \Gal (K/L)$ fixes $ \mathfrak{p}$ (since it is the only prime over $ \mathfrak{p}\cap L$), so $ \Gal (K/L)\subset D$, hence $ K^D\subset L$. $ \qedsymbol$

Thus $ p$ does not split in going from $ K^D$ to $ K$--it does some combination of ramifying and staying inert. To fill in more of the picture, the following proposition asserts that $ p$ splits completely and does not ramify in $ K^D/\mathbf{Q}$.

Proposition 14.1.4   Let $ L=K^D$ for our fixed prime $ p$ and Galois extension $ K/\mathbf{Q}$. Let $ e=e(L/\mathbf{Q}),f=f(L/\mathbf{Q}),g=g(L/\mathbf{Q})$ be for $ L/\mathbf{Q}$ and $ p$. Then $ e=f=1$ and $ g=[L:\mathbf{Q}]$, i.e., $ p$ does not ramify and splits completely in $ L$. Also $ f(K/\mathbf{Q})=f(K/L)$ and $ e(K/\mathbf{Q})=e(K/L)$.

Proof. As mentioned right after Definition 14.1.1, the orbit-stabilizer theorem implies that $ g(K/\mathbf{Q})=[G:D]$, and by Galois theory $ [G:D]=[L:\mathbf{Q}]$. Thus

$\displaystyle e(K/L)\cdot f(K/L)$ $\displaystyle = [K:L] =[K:\mathbf{Q}]/[L:\mathbf{Q}]$    
  $\displaystyle = \frac{e(K/\mathbf{Q})\cdot f(K/\mathbf{Q}) \cdot g(K/\mathbf{Q})}{[L:\mathbf{Q}]} = e(K/\mathbf{Q})\cdot f(K/\mathbf{Q}).$    

Now $ e(K/L)\leq e(K/\mathbf{Q})$ and $ f(K/L)\leq f(K/\mathbf{Q})$, so we must have $ e(K/L)=e(K/\mathbf{Q})$ and $ f(K/L)=f(K/\mathbf{Q})$. Since $ e(K/\mathbf{Q})=e(K/L)\cdot e(L/\mathbf{Q})$ and $ f(K/\mathbf{Q})=f(K/L)\cdot f(L/\mathbf{Q})$, the proposition follows. $ \qedsymbol$

William Stein 2004-05-06